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Difference between revisions of "2001 AMC 8 Problems/Problem 25"
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There are 24 four-digit whole numbers that use each of the four digits 2, 4, 5 and 7 exactly once. Only one of these four-digit numbers is a multiple of another one. Which of the following is it?
 
There are 24 four-digit whole numbers that use each of the four digits 2, 4, 5 and 7 exactly once. Only one of these four-digit numbers is a multiple of another one. Which of the following is it?
  
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<math>\text{(A)}\ 5724 \qquad \text{(B)}\ 7245 \qquad \text{(C)}\ 7254 \qquad \text{(D)}\ 7425 \qquad \text{(E)}\ 7542</math>
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<math>\textbf{(A)}\ 5724 \qquad \textbf{(B)}\ 7245 \qquad \textbf{(C)}\ 7254 \qquad \textbf{(D)}\ 7425 \qquad \textbf{(E)}\ 7542</math>
  
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==Solution==
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==Solution 1==
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We begin by narrowing down the possibilities. If the larger number were twice the smaller number, then the smallest possibility for the larger number is <math> 2457\times2=4914 </math>, since <math> 2457 </math> is the smallest number in the set. The largest possibility would have to be twice the largest number in the set such that when it is multiplied by <math> 2 </math>, it is less than or equal to <math> 7542 </math>, the largest number in the set. This happens to be <math> 2754\times2=5508 </math>. Therefore, the number would have to be between <math> 4914 </math> and <math> 5508 </math>, and also even. The only even numbers in the set and in this range are <math> 5472 </math> and <math> 5274 </math>. A quick check reveals that neither of these numbers is twice a number in the set. The number can't be quadruple or more another number in the set since <math> 2457\times4=9828 </math>, well past the range of the set. Therefore, the number must be triple another number in the set. The least possibility is <math> 2457\times3=7371 </math> and the greatest is <math> 2475\times3=7425 </math>, since any higher number in the set multiplied by <math> 3 </math> would be out of the range of the set. Reviewing, we find that the upper bound does in fact work, so the multiple is <math> 2475\times3=7425, \boxed{\text{D}} </math>
 
  
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There are only 5 options for the problem so we can just try them. It is easy since that we only need try to use <math>2</math>, <math>3</math> to divide them. Even <math>5</math> will leads to a solution start with <math>1</math> which we don't need.
  
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<math>5724/2=2862</math>, <math>5724/3=1908</math>. <math>7245/3=2415</math>. <math>7254/2=3612</math>, <math>7254/3=2418</math>. <math>7425/3=2475</math>. The answer is <math>\boxed{\textbf{(D)}}</math>. You can obtain the answer in only 6 calculations.
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Or, since the greatest number possible divided by the smallest number possible is slightly greater than 3, divide all of the choices by 2, then 3 and see if the resulting answer contains 2,4,5 and 7. Doing so, you find that <math> 7425/3 = 2475,  \boxed{\text{ D }}</math>
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~ M4ST3R0FM4TH
  
 
==See Also==
 
==See Also==
 
  {{AMC8 box|year=2001|num-b=24|after=Last<br />Question}}
 
  {{AMC8 box|year=2001|num-b=24|after=Last<br />Question}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:01, 26 July 2024

Problem

There are 24 four-digit whole numbers that use each of the four digits 2, 4, 5 and 7 exactly once. Only one of these four-digit numbers is a multiple of another one. Which of the following is it?

$\textbf{(A)}\ 5724 \qquad \textbf{(B)}\ 7245 \qquad \textbf{(C)}\ 7254 \qquad \textbf{(D)}\ 7425 \qquad \textbf{(E)}\ 7542$

Solution 1

There are only 5 options for the problem so we can just try them. It is easy since that we only need try to use $2$, $3$ to divide them. Even $5$ will leads to a solution start with $1$ which we don't need.

$5724/2=2862$, $5724/3=1908$. $7245/3=2415$. $7254/2=3612$, $7254/3=2418$. $7425/3=2475$. The answer is $\boxed{\textbf{(D)}}$. You can obtain the answer in only 6 calculations. ~ M4ST3R0FM4TH

See Also

2001 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last
Question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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